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Marizza181 [45]

3 years ago

6

374 exceeds three times a number by 74

2 answers:

mario62 [17]3 years ago

8 0

$3x = 374 + 74 \\ 3x = 448 \\ x = \frac{448}{3} \\ x = 149.333$

Assoli18 [71]3 years ago

3 0

$374>3x+74\\374=74>3x\\300>3x\ /:3\\100>x\\Ans.\ x$

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GarryVolchara [31]

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The unusual $X$ values for this model are: $X = 0, 1, 2, 7, 8$

Step-by-step explanation:

A binomial random variable $X$ represents the number of successes obtained in a repetition of $n$ Bernoulli-type trials with probability of success $p$ . In this particular case, $n = 8$ , and $p = 0.53$ , therefore, the model is ${8 \choose x} (0.53) ^ {x} (0.47)^{(8-x)}$ . So, you have:

$P (X = 0) = {8 \choose 0} (0.53) ^ {0} (0.47) ^ {8} = 0.0024$

$P (X = 1) = {8 \choose 1} (0.53) ^ {1} (0.47) ^ {7} = 0.0215$

$P (X = 2) = {8 \choose 2} (0.53)^2 (0.47)^6 = 0.0848$

$P (X = 3) = {8 \choose 3} (0.53) ^ {3} (0.47)^5 = 0.1912$

$P (X = 4) = {8 \choose 4} (0.53) ^ {4} (0.47)^4} = 0.2695$

$P (X = 5) = {8 \choose 5} (0.53) ^ {5} (0.47)^3 = 0.2431$

$P (X = 6) = {8 \choose 6} (0.53) ^ {6} (0.47)^2 = 0.1371$

$P (X = 7) = {8 \choose 7} (0.53) ^ {7} (0.47)^ {1} = 0.0442$

$P (X = 8) = {8 \choose 8} (0.53)^{8} (0.47)^{0} = 0.0062$

The unusual $X$ values for this model are: $X = 0, 1, 7, 8$

6 0

3 years ago

Read 2 more answers